Conception Aero Performances
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Transcript of Conception Aero Performances
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Introduction to Aircraft Design
Aircraft Design
Lecture 8:
Aircraft Performance
G. Dimitriadis
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Introduction to Aircraft Design
Design for Performance
The most important requirement for a newaircraft design is that it fulfills its mission
This is assured through performancecalculations at the design stage
As these calculations are carried out,important aircraft parameters are chosen: Size of wing
Engine
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Flight points
Performance calculations are crucial atseveral flight points. The most importantare:
Cruise
Take off
Climb
Landing
The first performance design analyses foran airliner are carried out for cruise
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Weight and drag
In order to choose parameters such as
engine and wing size, the aircrafts
weight and drag must be known.
Then the amount of lift and thrust
required can be determined
Of course, the weight and drag must be
calculated at several important points inthe flight envelope.
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Methodology
It is impossible to know the weight and
drag of an aircraft before it has even
been designed There are two possibilities:
Carry out detailed simulations at the
conceptual design stage; very costly
Use previous experience, statistical data,carry out industrial espionage etc
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Introduction to Aircraft Design
Statistics
There is an enormouswealth of data on civil
aircraft. There are
hundreds of types thatare very similar. They can
be used to extract somevery useful statistics.
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Weight guesstimates
The first important weight to calculate is thetake off weight, Wto.
It is usually expressed as:
Where Wpis the payload weight, Wfis the fuelweight, Wfixis the fixed empty weight (e.g.engine) and Wvaris the variable empty weight.
Notice that in this expression, We=Wfix+Wvaristhe total empty weight of the aircraft
Wto=
Wp +Wfix
1-Wvar
Wto
Wf
Wto
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Light aircraft (Wto
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Undercarriage drag correction
factor The undercarriage drag correction factor is used
both in the calculation of fuel weight and in thecalculation of the zero-lift drag (see later)
For a fully retractable landing gear that disappears
inside the aircraft lines, ruc=1. Otherwise:
ruc =
1.35 - for fixed gear without streamlined wheel fairings
1.25 - for fixed gear with streamlined wheel fairings
1.08 - main gear retracted in streamline fairings on the fuselage
1.03 - main gear retracted in engine nacelles
"
#
$$
%$$
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Introduction to Aircraft Design
Wheel Fairings
Cessna 180 with wheelfairing
Cessna 172 without wheelfairing
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Introduction to Aircraft Design
Transport Aircraft (Wto>5670Kg) Again, statistical studies show that
Where Wengis the engine weight and Weis calculated from a statistical graph. Allweights are in Kg.
The fuel weight is also calculated from astatistical graph
Wvar
Wto
= 0.2
Wfix =Weng + 500+ We
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We
calculation
lf = length of fuselagebf= width of fuselage
hf = height of fuselage
Use metric data
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Fuel weight -
Turboprops
Cp= Specific fuel
consumption forpropeller aircraft
Use metric data
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Fuel weight -
Turbojetsp : atmospheric pressure at cruiseconditions
M : Mach number at cruiseconditions
=T/T0, cruise temperature/
standard temperatureCT/: Corrected specific fuel
consumption at cruise conditionsa0: Speed of sound at sea level,
International Standard
Atmosphere: Mean skin friction coefficient
based on wetted area.
CF=
0.003 - for large, long- range transports
0.0035 - for small, short - range transports
0.004 - for business and executive jets
"
#$
%$
CF
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Introduction to Aircraft Design
Skin friction coefficient
An estimated of the drag force due to air frictionover the full surface of the aircraft (wettedarea).
It can be estimated using Prandtl-Schlichtingtheory as
WhereRecris the Reynolds number based oncruise flight conditions and the fuselage length
CF=
0.455
log10 Recr( )( )2.58
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More skin friction
Skin frictioncoefficients for
several aircraft
types.
rRe=C
F Re( ) /CF 10
8( )
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Caravelle type
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Boeing 707 type
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Boeing 727 type
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Avro RJ85/100 (Bae 146)
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Antonov An-72/74
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Turboprops
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Blended Wing Body
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Box wing
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Circular wing
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Drag Calculations
There are many sources of aircraft drag
They are usually summarized by the
airplane drag polar:
Where CD0is the drag that is
independent of lift and eis the Oswaldefficiency factor.
CD= C
D0
+
CL
2
eAR
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The drag polar
At high values of CLthe wing stalls
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Drag figures for different aircraft
Aircraft Type CD0 e
High-subsonic jet 0.014-0.020 0.75-0.85
Large turboprop 0.018-0.024 0.80-0.85
Twin-engine pistonaircraft
0.022-0.028 0.75-0.80
Single-engine pistonaircraft with fixed gear
0.020-0.030 0.75-0.80
Single-engine pistonaircraft with retractable
gear
0.025-0.040 0.65-0.75
Agricultural aircraftwithout spray system
0.060 0.65-0.75
Agricultural aircraft withspray system
0.070-0.080 0.65-0.75
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Compressibility drag
Compressibility effects increase drag
A simple way of including
compressibility effects inpreliminary drag calculations is
to add CDto CD0, where
CD=0.0005 for long-range
cruise conditions
CD=0.0020 for high-speedcruise conditions
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Cruise
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High speed performance
At cruise, the flight speed is constant.Therefore, T=D. This can be written as
Where is the cruise air density, Vis the cruise
airspeed and Sis the wing area. The drag coefficient is obtained from the drag
polar, using the fact thatL=W, i.e.
T= D =1
2V
2CDS
W = L =1
2V
2CLS
CL= W1
2V
2S
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Thrust to weight
The thrust to weight ratio is then
The thrust here is the installed thrust,
which is 4-8% lower than the un-
installed thrust
This equation can be used to choose anengine for the cruise condition
T
W=
1
2WV
2CD
0
+
CL
2
eAR
$
%&
'
()S=
V2CD
0
2W /S+
2W
eARV2S
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Minimum Thrust
The thrust-to-weight ratio can be minimized asa function of wing loading W/S.
Minimum thrust is required when
Where d1=0.008-0.010 for aircraft withretractable undercarriage. Therefore
W
S=
1
2V
2d1eAR
T
W"#$ %
&'min
= CD0 + d1
d1eAR
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Engine thrust
The thrust of an engine at the cruisecondition can be determined from:
Manufacturers data
Approximate relationship to the take offthrust:
where is the bypass ratio,Mis the cruise
Mach number and G=0.9 for low bypassengines and G=1.1 for high bypass
T
Tto
= 10.454 1+ ( )
1+ 0.75M+ 0.6 +
0.13
G
$%
&'M
2
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Wing loading and Mach
numberThe maximum wing loadingdepends uniquely on CLmax.
For an airliner, CLmaxdrops
to neary zero asM
approaches 1.
For supersonic aircraft,
CLmaxdrops in the transonic
region but not too much. It
then assumes a near
constant value for moderatesupersonic speeds
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The flight envelopeMMO,VMO=maximumoperating Mach number,
airspeed
MC, VC=design cruising
Mach number, airspeedMD,VD=design diving Machnumber, airspeed
VS=stalling airspeed
CAS=calibrated airspeed
(airspeed displayed on flight
instruments)TAS=true airspeed
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Determination of cruise
Mach number
MHS=high-speedMach number (for
high-speed cruise)
This calculation must
be repeated atseveral altitudes.
Each altitude
corresponds to a
different cruise Mach.
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Range performance
The range of an aircraft can be estimated
from the Brguet range equation:
This equation is applicable to cruise
conditions only, i.e.L/Dis the cruise lift-to-
drag ratio, Vis the cruise airspeed, Wiis
the weight of the aircraft at the start ofcruise and Wf is the cruise fuel weight
R =V
CT
L
D
ln W
i
WiWf
#
$%
&
'(
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Maximizing Range
The range equation can also be written as
WhereMis the cruise Mach number anda0is the speed of sound at sea level.
The range can be maximized either by
maximizingL/Dor by maximizingML/D.
Therefore:
R
a0
=
ML /D
CT /
ln W
i
WiW
f
$
%&
'
()
CL= C
D0eAR or CL = 1
3CD0eAR
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Lift to drag ratio
Example of lift to dragratio variation with lift,
angle of attack and the
deployment of slats and
flaps
The best L/D is obtained
for the clean configuration
Induced drag only
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Compressibility effect on
rangeWith compressibility thereis a global maximumML/D
Without compressibility there isno global maximum
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Some considerations
For long-haul aircraft the most importantconsideration is fuel efficiency
For short-haul aircraft the most importantconsideration is engine weight
There are some complications though: Cruise fuel is only part of the fuel weight
For short haul aircraft the engine thrust isfrequently determined by take off field length
Air traffic controllers decide the allowable cruise
altitudes
An aircraft can have more than one engine
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Reserve Fuel One definition of reserve fuel for international
airline operations (ATA 67): The airliner must carry enough reserve fuel to:
Continue flight for time equal to 10% of basic flighttime at normal cruise conditions
Execute missed approach and climbout at
destination airport Fly to alternate airport 370km distant
Hold at alternate airport for 30 minutes at 457mabove the ground
Descend and land at alternate airport
One approximate calculation:W
fres/W
to=0.18C
T / AR
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Range for propeller aircraft
The Brguet range equation for propelleraircraft is
Where pis the propeller efficiency and Cpis the specific fuel consumption.
The range can be maximized by: Minimizing airplane drag
Minimizing engine power
R =p
CP
L
Dln
Wi
WiW
f
$
%&
'
()
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Payload-range diagram
Two cases: high-speed cruise, long-rangecruise. Example: A-300B
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Take off
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Take Off Performance
T
D
L
R
W
V
x
V1VLOFV=0
V2
h=35ft
Ground Run Rotation Transition Climb
xg xa
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Take Off Description
Take off starts at time t0, with airspeed V0andthe runway may have an angle to horizontalof .
Lift off occurs at time tg, after a distance ofxg,usually at speed VLOF.
Take off is completed when the aircraft hasreached sufficient height to clear an obstacle35ft high (50ft for military aircraft)
Finally, the climb out phase takes the aircraftto 500ft at the climb throttle setting.
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Ground Run
The ground run can either start at V0=0.
The angle of attack is defined with respect tothe thrust line. For aircraft with a tricycle landing gear the angle of
attack is nearly zero. At a sufficient speed thenose is lifted up a little bit to achieve an optimumangle of attack.
For aircraft with a tailwheel the angle of attack isvery high. At an airspeed where the controlsurfaces are effective the angle of attack is
decreased to decrease the drag and increaseacceleration.
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Lift off
In principle an aircraft can lift off once it hasreached its stall speed, Vstall.
At this airspeed it can increase its angle of
attack max, with a resulting CLmax. This liftcoefficient will generally satisfyL>W.
However, for safety reasons, the lift off speedis defined as VLOF=k1Vstallwhere
k1=1.10is an indicative value
In general k1varies with type of aircraft
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Rotation and Transition
After lift off the aircraft rotates to deflect thevelocity from nearly horizontal to a fewdegrees upward.
The rotation lasts usually for a couple of
seconds. The transition stage follows, at the end of
which the aircraft is required to clear ahypothetical obstacle.
The speed at the clearance of the obstacle isusually V2=k2Vstallwhere k2=1.2.
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Take off details
V1is called the decision speed for multi-engined aircraft. Below this speed, if one engine fails, the take off is
aborted.
Above this speed, if one engine fails, the take off
is continued. Take off field length: Distance from rest to
obstacle clearance
The climb stage follows the transition stage.
The take off phase is usually considered to be
complete at around 500ft once the flaps havebeen retracted.
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Equations of motion
The total force applied to the aircraft parallelto the runway is:
The acceleration is equal to a=dV/dt. Considering that the distancexis given by dx/
dt=V,
dx
dt
dt
dV=
V
a so that
xg=
V
a0
VLOF
dV
F = TR D = ma
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Friction force
In order to have a friction force, it must
be thatR>0, i.e.
Where it is assumed that the runways
inclination is small. The total force can
be written as
R =W L =W1
2 V
2SC
L
ma = TW 1
2V
2S C
DC
L( )
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Friction Coefficients
Runway type
Concrete, asphalt 0.02
Hard turf 0.04
Field with short grass 0.05
Field with long grass 0.10
Soft field, sand 0.10-0.30
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Approximate solution (1)
An approximate solution for the ground
run can be obtained as
xg V
LOF
2/2g
TW
to
$
T = thrust at VLOF
/ 2 0.755 +
4 + T
to
CL = eAR
$ = + 0.72CD0
CLmax
where
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Approximate solution (2)
An approximate solution for the air run
can be obtained as
The airspeed at the takeoff obstacle is
xa V
LOF
2
g 2+
h
LOF
where
LOF
=
TD
W
$
%&
'
()LOF
0.9T
Wto
0.3
AR
V2=V
LOF 1+
LOF 2
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Climb
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Climb performance
Operational requirements, e.g.
Rate of climb at sea level
Service ceiling altitude for a maximum rate
of climb of 0.5m/sAirworthiness requirements
Minimum climb gradient in takeoff, en
route, landing
Rate if climb at a specified altitude with oneengine inoperative
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Climb description
Climb follows immediately after take off andits objective is to bring the aircraft to cruisingheight
In general, climb is performed in a vertical
plane, i.e. no turning during climb.
Climb takes up a relatively short percentageof flight time so it can be assumed that theaircrafts weight remains constant
Variations in speed and flight path areconsidered small, i.e. constant speed climb
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Climb diagram
L
D
VT
W
x
z
Notice that thethrust line is not
necessarily
parallel to the flight
path
Also note that, for
constant speed V,
the aircraft cannot
be accelerating
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Thrust and Lift equations
Assume that =0. Then, write theequilibrium equations on the body axes:
As mentioned before, the drag can be
expressed in terms of lift using the dragpolar
T =1
2
SV2CD +Wsin
Wcos =1
2SV
2CL
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Thrust and Power
It is customary to express the thrust equationin terms of power by multiplying it by V:
Define
WherePais the power available for climb,Pris the power required for level flight at thesame airspeed and Vzis the rate of climb.
Then
TV =1
2SV
3CD +WVsin
Pa = TV, P
r =
12SV
3C
D, V
z = Vsin
Vz =
Pa
Pr
W and sin =
Pa
Pr
VW
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Climb of a jet aircraft
For jet aircraft the amount of power available
for climb varies linearly with airspeed.
The power required for level flight varies non-
linearly with airspeed.
There is an optimum airspeed for maximum
rate of climb which maximizesPa-Pr.
This airspeed is not necessarily at the
airspeed yielding the minimum value ofPr.
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Optimum airspeedThe optimum climb rate isobtained when the powerdifference Pismaximized.
It occurs at the pointwhere the tangent to curve
Pris parallel toPa.At airspeeds below V2andabove V1the aircraftcannot climb; the powerrequired for level flight ishigher than the poweravailable for climb.
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Rate of climb
The equation for the rate of climb can bewritten as
Where a parabolic drag polar was substitutedfor CDand the rate of climb was assumedsmall.
The maximum climb rate is obtained whenVz/dV=0, which gives:
Vz =
Pa P
r
W=
1
WTV
1
2SV
3CD0
2KW
2
SV
#
$%&
'(
3SCD0
WV
4
2T
WV
2
4KW
S= 0
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Solution for maximum climb
rate The quartic equation has four double roots. They
depend on the discriminant
The solutions are
AsDis always greater than 2T/W, the only positivesolution is
D =2T
W
!"#
$%&
2
43SC
D0
W
!
"#$
%&
4KW
S
!
"#$
%& = 2
T
W
!"#
$%&
2
+
3
Emax
2
V2=
2T
W D
!"#
$%&
/ 23SC
D0
W
!
"#$
%&
V2 = W3SC
D0
TW
+ TW
"#$
%&'
2
+
3
Emax
2
"
#$$
%
&''
where Emax
=
CL
CD
"
#$
%
&'max
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Maximum climb rate
Substituting this value of airspeed in the climbrate equation we get
Where =EmaxT/W. Therefore, the maximum climb rate depends on:
Thrust available
Weight
Altitude
Wing surface
Vzmax
V=
1
3Emax
+ 2+ 3( ) 3
Emax
+ 2+ 3
( )
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Climb rate requirements
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Flight Dynamics 08
Greg Dimitriadis
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Turning
The most general turn is a turnaccompanied by a change in height.
All the turn angles of the aircraft are
involved, roll, pitch and yaw In these lectures we will only cover
turns in a horizontal plane, i.e. nochange of height
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Turn diagrams
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Equations of motion
The equations of motion for the most
general case are:dx
dt= Vcoscos
dy
dt= Vcossin
dz
dt= Vsin
dV
dt=g
T
Wcos
D
Wsin
%&'
()*
d
dt=g
V
L
W+T
Wsin
%&'
()*cos cos%
&'()*
d
dt=
g
V
L
W+
T
Wsin
%&'
()*sin
cos
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Constant speed horizontal
turn For a horizontal turn, =0.
For constant speed, dV/dt=0.
Also assume that there is no anglebetween the thrust line and theprojection of the flight path on theaircrafts plane of symmetry, i.e. =0.
The equations of motion become
Lcos =W, T =D
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Load factor
The load factor is quite important for turns. Ifthe load factor is too great the pilot can beharmed or the aircrafts structure can bedamaged.
The load factor is defined as n=L/W.
From the equations of motion:
During a turn, the lift must balance not onlythe weight but also the centrifugal force. This
means that the loads acting on the aircraft arefar more important.
n = 1 / cos
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Example
Assuming that the maximum load factor
allowed for an aircraft is nmax, calculate
the thrust required to perform a
horizontal turn of radiusRat a constant
airspeed V.
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Solution (1)
We start with determining the load
factor. From the turning diagram, it is
seen that:
Now calculate the required lift. The lift
equation can be written as:
nW = mg( )2 + mV2
R!"#
$%&
2
nW =1
2SV
2CL
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Solution (2)
The drag can be obtained, as usual, by
a drag polar, i.e.
The thrust is then given by
CD = CD0 + KCL2
= CD0 + K
2nW
SV2
"
#$%
&'
2
T = 12
SV2CD = 1
2SV2 C
D0
+ K 2nWSV
2"#$
%&'
2"
#$
%
&'
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Solution (3)
This is the thrust required to perform the
required turn.
It must now be verified that the radius R
will correspond to a load factor lower
than nmax.
mg( )2 +mV
2
R
!
"#$
%&
2
W < nmax
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Introduction to Aircraft Design
The turning rate
For transport aircraft, the control ofturning rate to complete a turn in aprescribed time is important.
For military aircraft, a high turning rate
is vital (literally). The turning rate is defined as d/dt. The
last of the equations of motion gives (for=0, =0)
ddt
=gV
LW
sin
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Introduction to Aircraft Design
The turning rate (2)
Using the lift equation, the turning rate
can be calculated as
If we express the airspeed in terms of
the load factor and the lift coefficient we
get
d
dt
=
g
V
tan =g n 1
V
d
dt=g SCL
2Wn 1
n$%&
'()
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Introduction to Aircraft Design
Maximum turning rate
The equation for the turning rate contains thelift coefficient and the load factor
The lift coefficient can never be higher than
CLmax
.
The load factor can never be higher than nmax.
Therefore the maximum turning rate is
obtained by
ddt
"#$ %&'max
=g SCLmax
2Wnmax
1
nmax
"#$
%&'
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Introduction to Aircraft Design
The turning radius
The turning radius can also be
expressed as
Expressing the airspeed in terms of the
load factor and lift coefficient we get:
R =V
2
g tan=
V2
g n
21
R =2W
gSCL"#$
%&'
nn2 1
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Introduction to Aircraft Design
Thrust required for turning
The thrust equations can be written
directly as:
If it is assumed that the lift to drag ratio
is constant, then the thrust required for
turning is directly proportional to the
load factor.
Tr = nW
CD
CL
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Introduction to Aircraft Design
Power required for turning Multiplying the thrust equation by Vwe
obtained the power required for turning, i.e.
Using the drag polar to replace the dragcoefficient and the drag equation to write the
lift coefficient in terms of nand Vwe get:
Finally, replacing for nin this expression:
P =1
2SV
3C
D
P =1
2SV
3CD
0
+
2Kn2W
2
SV
P =1
2SV3CD0 +
2K
!
2
W
2
Vg2S
+
2KW
2
SV
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Introduction to Aircraft Design
Landing
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Introduction to Aircraft Design
Landing
The landing is performed in two phases:
Approach over a hypothetical obstacle to touch-
down
Ground run to a full stop Before reaching the obstacle the aircraft
makes an approach along the axis of the
runway with a glide angle between -2.5 and
-3.5 degrees. The approach speed is V2, i.e. 1.2Vstall.
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Introduction to Aircraft Design
Landing diagram
hrVTD V=0
V235ft
Approach Rotation (flare) Ground run Stop
V2
V2
Deceleration
x3= approach distance
x2= rotation distance
x1= deceleration distance
xg= ground run distance
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Introduction to Aircraft Design
Approach
The equations of motion are as for thedescent. For a generalized approach, sincethe angle is small:
whereE=CL/CD. The lift coefficient is fixed bythe second equation. The drag coefficient canbe obtained from the drag polar.
Therefore, for a chosen approach angle, the
first equation will yield the correct thrustsetting.
=1
E
T
W, and
1
2V
2
2CL =W
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Introduction to Aircraft Design
Approach (2)
If the height of the hypothetical object is
hoband the rotation height is hr, then the
approach distancex3is given by
And the approach time is
x3 =
hobh
r
tan
t3 = x
3
V2cos
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Introduction to Aircraft Design
Rotation
Rotation is a circular arc with radiusR. As in the take-off case, we have
Where nis the load factor.
The distance traveled during rotation is
And the time taken is
R =V2
2
g(n 1)
x2 = Rsin
t2 =V2
g(n 1)
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Introduction to Aircraft Design
Ground run
Once the aircraft has touched down theairspeed drops from VTDto 0.
The distance of the ground run can beapproximated by
Where is the mean deceleration:
xg =VTD2/2a
a
a =
0.30 0.35 for light aircraft with simple brakes
0.35 - 0.45 for turboprop aircraft without reverse propeller thrust
0.40 0.50 for jets with spoilers, anti - skid devices, speed brakes
0.50 - 0.60 as above, with nosewheel breaks
#
$
%%
&%
%
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Introduction to Aircraft Design
Parametric design
Design for performance is above all aprocess of optimization.
The aim is to satisfy or exceed allperformance requirements by finding theoptimal combination of parameters
The parameters are: Poweplant: Takeoff thrust, number of engines,
engine type of configuration
Wing: Wing area, Aspect Ratio, high-liftdevices
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Flow diagram